a body is projected vertically upwards with a velocity of 20m/s It reaches the maximum vertical height h in time t in time 1/2 the distance of the body form its highest position is
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Answered by
2
Explanation:
here
initial speed(u)=20m/s
therefore
final velocity (v)=0m
g=9.8m/s^2
time=t
according to first equation of motion
u=v-gt
20=0-9.8×t
t=2.1s
max height=u^2/2g
=400/19.6
=20.4m
height attain at t=1/2s is
s=ut-1/2gt^2
s=20×1.05-4.9×1.1
s=21-5.39
s=15.61
therefore distance from highest point from t=1/2 is
max height-given height
20.4-15.61= 4.79m
Answered by
0
Explanation:
here
initial speed(u)=20m/s
therefore
final velocity (v)=0m
g=9.8m/s^2
time=t
according to first equation of motion
u=v-gt
20=0-9.8×t
t=2.1s
max height=u^2/2g
=400/19.6
=20.4m
height attain at t=1/2s is
s=ut-1/2gt^2
s=20×1.05-4.9×1.1
s=21-5.39
s=15.61
therefore distance from highest point from t=1/2 is
max height-given height
20.4-15.61= 4.79m
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